Chapter 11.2, Problem 43WE

a.

To determine

To confirm the given fact.

Explanation of Solution

Given:

The given fact is $\frac{n}{99}=n(0.\overline{01})$ for $1\le n<100$

Calculation:

The given fraction $\frac{8}{99}$ can be written as

  $\begin{array}{l}\frac{8}{99}=0.0808080808\\ \frac{8}{99}=0.\overline{08}\\ \frac{8}{99}=8(0.\overline{01})\end{array}$

Similarly, the given fraction $\frac{32}{99}$ can be written as

  $\begin{array}{l}\frac{32}{99}=\mathrm{0.323232.....}\\ \frac{32}{99}=0.\overline{32}\\ \frac{32}{99}=32(0.\overline{01})\end{array}$

And $\frac{87}{99}=87(0.\overline{01})$

Thus verified.

b.

To determine

To show that $0.\overline{9}=1$

Explanation of Solution

The $1$ can be written as $1=\frac{99}{99}$

The given fraction $\frac{99}{99}$ can be written as

  $\begin{array}{l}\frac{99}{99}=\mathrm{0.999999....}\\ \frac{99}{99}=0.\overline{9}\\ \frac{99}{99}=\frac{9}{9}=1\end{array}$

c.

To determine

To show that $0.\overline{9}=1$ using method of example 4.

Explanation of Solution

The given fraction $0.\overline{9}$ can be written as

  $0.\overline{9}=\frac{9}{10-1}=\frac{9}{9}=1$